$ f(x)=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{{{x}^{2n+1}}}{\left( 2n \right)!}}$ Find the power series of $f''(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+1 \right)\left( 2n \right){{x}^{2n}}}{\left( 2n \right)!}}$ (Choice B) B $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\dfrac{\left( 2n+1 \right)\left( 2n \right){{x}^{2n-1}}}{\left( 2n \right)!}}$ (Choice C) C $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+1 \right)\left( 2n \right){{x}^{2n-1}}}{\left( 2n-1 \right)!}}$ (Choice D) D $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+1 \right){{x}^{2n-1}}}{\left( 2n-1 \right)!}}$ (Choice E) E $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+1 \right)\left( 2n \right){{x}^{2n}}}{\left( 2n-1 \right)!}}$
Answer: $\begin{aligned} &\phantom{=} {f}\,'\left( x \right) \\\\ &=1-\dfrac{3{{x}^{2}}}{2!}+...+{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+1 \right){{x}^{2n}}}{\left( 2n \right)!}+... \\\\ &=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+1 \right){{x}^{2n}}}{\left( 2n \right)!}} \\\\ \\\\ &\phantom{=} {f}\,''\left( x \right) \\\\ &=-\dfrac{3\cdot 2x}{2!}+...+{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+1 \right)\left( 2n \right){{x}^{2n-1}}}{\left( 2n \right)!}+... \\\\ &=\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+1 \right)\left( 2n \right){{x}^{2n-1}}}{\left( 2n \right)!}} \end{aligned}$ Now note that $\dfrac{2n}{\left( 2n \right)!}=\dfrac{1}{\left( 2n-1 \right)!}$. Hence, ${f}\ ''\left( x \right)=\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{\left( 2n+1 \right){{x}^{2n-1}}}{\left( 2n-1 \right)!}}$